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4 years ago

How do I solve this?

Mathematics

1 answer:

devlian [24]4 years ago

5 0

Answer: $3.00

Well, first, you'll need to add $1.50 four times. You'll have $6.00. So, two pairs of scissors cost $6.00. Then, you divide $6.00 by 2.

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N equation of a line through (0, 2) which is perpendicular to the line y=3x+3y=3x+3 has slope

zhannawk [14.2K]

Something's not quite right here. What do you mean by "N?"

I could and will make assumptions regarding what you meant. (Feel free to correct me if need be.)

Find the equation of a line through (0,2) which is perpendicular to the line y=3x+3. To find the slope of this perpendicular line, take the negative reciprocal of 3 (the slope of the given line). It is -1/3.

Using the point-slope form of the equation of a straight line:

y = (-1/3)x + 2 (answer)

7 0

3 years ago

F ∠A is 95° and ∠B is 105°, what is ∠C? A) 75° B) 85° C) 95° D) 105°

Jobisdone [24]

The question is missing important data. The quadrilateral having vertices A, B, C and D is a cyclic quadrilateral.

Answer:

B) 85°

Step-by-step explanation:

Given:

A cyclic quadrilateral with ∠A is 95° and ∠B is 105°.

For a cyclic quadrilateral, the sum of the opposite interior angles is equal to 180°

Therefore, sum of angles A and C is 180°.

$\angle A + \angle C=180\°\\95\°+\angle C=180\°\\\angle C=180-95\\\angle C =85\°$

Therefore, the correct option is option B) 85°

6 0

4 years ago

Read 2 more answers

The amphitheater has two types of

MakcuM [25]

Answer:

20,000 reserved seat tickets; 0 lawn seat tickets.

Step-by-step explanation:

There is no limit on the number of reserved seats that can/must be sold. Selling the maximum possible number of seats that give the larger profit will maximize profit.

6 0

4 years ago

B+7 ÷ 2a for a = -4 b= -6

Evgen [1.6K]

-6+7 / 2(-4)=

-8

PEMDAS

6 0

3 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and asymptotes

$y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x$

b) The length of the transverse axis is given by $2a$ . Therefore, the lenght is 6.

c) See below.

4 0

4 years ago